hdu 2389 Rain on your Parade （二分匹配 Hc 算法）

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题目链锁 http://acm.hdu.edu.cn/showproblem.php?pid=2389
先生成最短增广路，在此基础进行增广，复杂度sqrt(n)*e
#include<iostream>#include<cstring>#include<string>#include<cstdio>#include<stdio.h>#include<algorithm>#include<cmath>#include<set>#include<map>#include<queue>#include<vector>using namespace std;#define inf 0x3f3f3f3f#define eps 1e-9#define mod 1000000007#define FOR(i,s,t) for(int i = s; i < t; ++i )#define REP(i,s,t) for( int i = s; i <= t; ++i )#define LL long long#define ULL unsigned long long#define pii pair<int,int>#define MP make_pair#define lson id << 1 , l , m#define rson id << 1 | 1 , m + 1 , r #define maxn ( 3000+10 )#define maxe ( 50000+10 )struct node {    int x, y;    int dis ( node ot ) {        return ( x - ot.x ) * ( x - ot.x ) + ( y - ot.y ) * ( y - ot.y );    }}g[maxn], u[maxn];int sp[maxn];int G[maxn][maxn];int mx[maxn], my[maxn], dx[maxn], dy[maxn];int dis, nx, ny;int Q[maxn*2];bool vis[maxn];bool search() {    memset( dx, -1, sizeof( dx ) );    memset( dy, -1, sizeof( dy ) );    dis = inf;    int head = 0, tail = 0;    for( int i = 0; i < nx; ++i ) {        if( mx[i] == -1 ) Q[tail ++] = i, dx[i] = 0;    }    while( head < tail ) {        int u = Q[head++];        if( dx[u] > dis ) break;        for( int v = 0; v < ny; ++v ) if( G[u][v] && dy[v] == -1 ) {            dy[v] = dx[u] + 1;            if( my[v] == -1 ) dis = dy[v];            else {                dx[my[v]] = dy[v] + 1;                Q[tail++] = my[v];            }        }    }    return dis != inf;}bool dfs ( int u ) {    for( int i = 0; i < ny; ++i ) if( G[u][i] && !vis[i] && dy[i] == dx[u] + 1 ) {        vis[i] =1;        if( my[i] != -1 && dis == dy[i] ) continue;        if( my[i] == -1 || dfs( my[i] ) ) {            my[i] = u;            mx[u] = i;            return 1;        }    }    return 0;}int match() {    int res = 0;    memset( mx, -1, sizeof( mx) );    memset( my, -1, sizeof( my ) );    while( search() ) {        memset( vis, 0, sizeof( vis ) );        for( int i= 0; i < nx; ++i )            if( mx[i] == -1 && dfs( i ) ) ++res;    }    return res;}int main () {    int T;    scanf("%d", &T );    int cas = 1;    for( cas = 1; cas <= T; ++cas ) {        int n, m, t;        scanf("%d%d", &t, &m ) ;        for( int i = 0; i < m; ++i ) {            scanf("%d%d%d", &g[i].x, &g[i].y, &sp[i] );        }        scanf("%d", &n );        for( int i = 0; i < n; ++i ) {            scanf("%d%d", &u[i].x, &u[i].y );        }        memset( G, 0, sizeof( G ) );        for( int i = 0; i < m; ++i ) {            for( int j = 0; j < n; ++j ) {                if( g[i].dis( u[j] ) <= ( sp[i] * t ) * ( sp[i] * t ) )                    G[i][j] = 1;            }        }        nx = m, ny = n;        printf("Scenario #%d:\n", cas );        printf("%d\n\n", match() );    }}
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